关于C++中_finite()函数的说明

 The function int _finite(double x) returns 1 (true) if x is an ordinary number and 0 (false) if x is either infinite or not-a-number (NaN).

msdn中如下描述中INF和NaN  
  _finite   returns   a   nonzero   value   (TRUE)   if   its   argument   x   is   not   infinite,   that   is,   if   –INF   <   x   <   +INF.   It   returns   0   (FALSE)   if   the   argument   is   infinite   or   a   NaN.

其中INF表示无穷大（浮点数有上溢出），

NaN标示not a number.